Geoscience Reference
In-Depth Information
V
1
/
T
1
=
V
2
/
T
2
(15.5)
In this equation, the subscript 1 indicates the initial volume and temperature, and the subscript 2
indicates the volume and temperature after the change. Temperature, incidentally, needs to be given
in Kelvin, not Celsius, because if we have a temperature below 0°C the calculation works out so the
volume of the gas is negative, and we can't have a negative volume.
■
EXAMPLE 15.2
Problem
: If we have 2 L of methane gas at a temperature of 40°C, what will be the volume be if we
heat the gas to 80°C?
Solution
: The first thing we have to do is convert the temperatures to Kelvin, because Celsius cannot
be used in this equation:
40°C + 273 = 313 K
80°C + 273 = 353 K
We are now ready to insert these numbers into the equation:
V
1
/
T
1
=
V
2
/
T
2
2 L/313 K =
x
L/353 K
x
= 2.26 L
15.5.2.1.3 Gay-Lussac's Law
Gay-Lussac (1802) found that all gases increase in volume for each 1°C rise in temperature, and this
increase is equal to approximately 1/273.15 of the volume of the gas at 0°C.
P
1
/
T
1
=
P
2
/
T
2
(15.6)
If we increase the temperature of a container with fixed volume, the pressure inside the container
will increase.
15.5.2.1.4 Combined Gas Law
This law
combines the parameters of the preceding equations, forming
(
P
1
×
V
1
)/
T
1
= (
P
2
×
V
2
)/
T
2
(15.7)
The advantage of this equation is that whenever we are changing the conditions of pressure, volume,
and/or temperature for a gas, we just insert the numbers into this equation.
■
EXAMPLE 15.3
Problem:
If we have 2 L of a gas at a temperature of 420 K and decrease the temperature to 350 K,
what will the new volume of the gas be?
Solution
: To solve this problem, we use the combined gas law to find the answer. Because pressure
was never mentioned in this problem, we ignore it. As a result, the equation will be
V
1
/
T
1
=
V
2
/
T
2
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